Solve x/x+4-2/x-1=

$$\frac { x } { x + 4 } - \frac { 2 } { x - 1 } =$$

Evaluate

$$\frac{x^{2}-3x-8}{\left(x-1\right)\left(x+4\right)}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x+4 and x-1 is \left(x-1\right)\left(x+4\right). Multiply \frac{x}{x+4} times \frac{x-1}{x-1}. Multiply \frac{2}{x-1} times \frac{x+4}{x+4}.

$$\frac{x\left(x-1\right)}{\left(x-1\right)\left(x+4\right)}-\frac{2\left(x+4\right)}{\left(x-1\right)\left(x+4\right)}$$

Since \frac{x\left(x-1\right)}{\left(x-1\right)\left(x+4\right)} and \frac{2\left(x+4\right)}{\left(x-1\right)\left(x+4\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(x-1\right)-2\left(x+4\right)}{\left(x-1\right)\left(x+4\right)}$$

Do the multiplications in x\left(x-1\right)-2\left(x+4\right).

$$\frac{x^{2}-x-2x-8}{\left(x-1\right)\left(x+4\right)}$$

Combine like terms in x^{2}-x-2x-8.

$$\frac{x^{2}-3x-8}{\left(x-1\right)\left(x+4\right)}$$

Expand \left(x-1\right)\left(x+4\right).

$$\frac{x^{2}-3x-8}{x^{2}+3x-4}$$

Differentiate w.r.t. x

$$\frac{2\left(3x^{2}+4x+18\right)}{\left(\left(x-1\right)\left(x+4\right)\right)^{2}}$$